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Numbers divisible by 50 not divisible by 100 would be divisible by \(2^1\) and \(5^2\), but power of 2 cannot be more than 1 as anything including \(2^2\) and \(5^2\) would be divisible by 100. So only 1 power of 2 allowed(2^1) and 5 powers of 5(\(5^2\),\(5^3\),\(5^4\),\(5^5\),\(5^6\))=1*6*5*9(1 powers of 2)*(6 powers of 3)*(5 powers of 5)*(9 powers of 7)=270

Re: If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by [#permalink]
06 May 2013, 09:12

4

This post receivedKUDOS

Expert's post

GMATtracted wrote:

If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by 50 but NOT by 100?

A. 240B. 345C. 270D. 120E. None of these

To be divisible by 50, the factor must have \(2*5^2\) and to be not divisible by 100, it must NOT have \(2^2*5^2\). Hence the only constraints are on the power of 2 (which must be 1) and the power of 5 (which must be greater than or equal to 2)

The other prime factors can appear in any way in the factor. So number of factors = 1*(6)*(5)*(9) = 2701 - because 2 can have a power in only one way6 - because 3 can have 6 different powers (0/1/2/3/4/5)5 - because 5 can have 5 different powers (2/3/4/5/6)9 - because 7 can have 9 different powers (0/1/2/3/4/5/6/7/8)

Re: If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by [#permalink]
21 May 2013, 21:42

VeritasPrepKarishma wrote:

GMATtracted wrote:

If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by 50 but NOT by 100?

A. 240B. 345C. 270D. 120E. None of these

To be divisible by 50, the factor must have \(2*5^2\) and to be not divisible by 100, it must NOT have \(2^2*5^2\). Hence the only constraints are on the power of 2 (which must be 1) and the power of 5 (which must be greater than or equal to 2)

The other prime factors can appear in any way in the factor. So number of factors = 1*(6)*(5)*(9) = 2701 - because 2 can have a power in only one way6 - because 3 can have 6 different powers (0/1/2/3/4/5)5 - because 5 can have 5 different powers (2/3/4/5/6)9 - because 7 can have 9 different powers (0/1/2/3/4/5/6/7/8)

Re: If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by [#permalink]
21 May 2013, 22:50

Expert's post

sharmila79 wrote:

VeritasPrepKarishma wrote:

GMATtracted wrote:

If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by 50 but NOT by 100?

A. 240B. 345C. 270D. 120E. None of these

To be divisible by 50, the factor must have \(2*5^2\) and to be not divisible by 100, it must NOT have \(2^2*5^2\). Hence the only constraints are on the power of 2 (which must be 1) and the power of 5 (which must be greater than or equal to 2)

The other prime factors can appear in any way in the factor. So number of factors = 1*(6)*(5)*(9) = 2701 - because 2 can have a power in only one way6 - because 3 can have 6 different powers (0/1/2/3/4/5)5 - because 5 can have 5 different powers (2/3/4/5/6)9 - because 7 can have 9 different powers (0/1/2/3/4/5/6/7/8)

Re: If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by [#permalink]
29 Jun 2014, 06:16

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